1
On the study of various equations concerning the Isoperimetric Theorems.
Possible mathematical connections with some sectors of Number Theory, String
Theory and Eternal Inflation model. III
Michele Nardelli
1
, Antonio Nardelli
2
Abstract
In this paper (part III), we analyze various equations concerning the Isoperimetric
Theorems. We describe the new possible mathematical connections with some sectors
of Number Theory, String Theory and Eternal Inflation model
1
M.Nardelli studied at Dipartimento di Scienze della Terra Università degli Studi di Napoli Federico II,
Largo S. Marcellino, 10 - 80138 Napoli, Dipartimento di Matematica ed Applicazioni “R. Caccioppoli” -
Università degli Studi di Napoli “Federico II” – Polo delle Scienze e delle Tecnologie Monte S. Angelo, Via
Cintia (Fuorigrotta), 80126 Napoli, Italy
2
A. Nardelli studies at the Università degli Studi di Napoli Federico II - Dipartimento di Studi Umanistici –
Sezione Filosofia - scholar of Theoretical Philosophy
3
Introduction
In 1983, it was shown that inflation could be eternal, leading to a multiverse in which
space is broken up into bubbles or patches whose properties differ from patch to
patch spanning all physical possibilities.
When the false vacuum decays, the lower-energy true vacuum forms through a
process known as bubble nucleation. In this process, instanton effects cause a
bubble containing the true vacuum to appear. The walls of the bubble (or domain
walls) have a positive surface tension, as energy is expended as the fields roll over
the potential barrier to the true vacuum.
In mathematics, a ball is the space bounded by a sphere. It may be a closed ball
(including the boundary points that constitute the sphere) or an open ball (excluding
them). (From Wikipedia)
We propose that some equations concerning the “balls”, thus various sectors and
theorems of Geometric Measure Theory, can be related with several parameters of
some cosmological models as the “Multiverse” and the “Eternal Inflation” linked to
it, which provides that space is divided into bubbles or patches whose properties
differ from patch to patch and spanning all physical possibilities.
4
From:
Isoperimetry and Stability Properties of Balls with Respect to Nonlocal Energies
A. Figalli, N. Fusco, F. Maggi, V. Millot, M. Morini - Commun. Math. Phys Digital
Object Identifier (DOI) 10.1007/s00220-014-2244-1
We have that:
5
Now, we analyze the eqs. (7.2), (7.3), (7.4), (7.5) and (7.6)
For n = 3, k = 2, s = α = 1/2 , from
we obtain:
2(2+3-2) = 6
6
From:
we obtain:
(2^(1-0.5)*Pi)/(1+1/2)* gamma(1/2*(1-
1/2))/gamma(1/2*(3+1/2))*(((((gamma(2+1/2(3+1/2)))/gamma(2+1/2(3-2-1/2))-
gamma(1/2(3+1/2))/gamma(1/2(3-2-1/2)))))
Input
Result
42.6517…
The study of this function provides the following representations:
Alternative representations
7
Series representations
8
Integral representations
9
For n = 3, k = 2, α = 1/2
From:
we obtain:
10
(2^(1+0.5)*Pi)/(1-1/2)* gamma(1/2*(1+1/2))/gamma(1/2*(3-
1/2))*(((((gamma(2+1/2(3-1/2)))/gamma(2+1/2(3-2+1/2))-gamma(1/2(3-
1/2))/gamma(1/2(3-2+1/2)))))
Input
Result
20.3103…
The study of this function provides the following representations:
Alternative representations
11
Series representations
12
Integral representations
13
For n = 3, k = 2, α = 2
From:
we obtain:
(2^2*Pi)*gamma(1/2*(2-1))/gamma(1/2*(3-2))*(((((gamma(1/2(3-2)))/gamma(1/2(3-
2+2))-gamma(2+1/2(3-2))/gamma(2+1/2(3-2+2)))))
14
Input
Exact result
Decimal approximation
20.106192982….
The study of this function provides the following representations:
Property
Alternative representations
15
Series representations
Integral representations
16
For n = 3
From:
we obtain:
(2^2*Pi)/gamma(1/2(3-1))*(((((digamma(2+1)))/gamma(2+1/2(3-1))-
digamma(1/2(3-1))/gamma(1/2(3-1)))))
Input
Exact result
17
Decimal approximation
13.051530945…
The study of this function provides the following representations:
Alternate forms
Expanded form
Alternative representations
18
Series representations
Integral representations
19
From the sum of the previous results/expressions, we obtain:
6+42.6517+20.3103+((32π)/5)+[((((4(1/2(3/2-0.5772156649)+0.5772156649)π))))]
Input interpretation
Result
102.120….
The study of this function provides the following representations:
Alternative representations
20
Series representations
21
Integral representations
From which, we obtain:
17((6+42.6517+20.3103+((32π)/5)+[((((4(1/2(3/2-
0.5772156649)+0.5772156649)π))))]))-e*Pi+φ
Input interpretation
Result
1729.11….
22
This result is very near to the mass of candidate glueball f
0
(1710) scalar meson.
Furthermore, 1728 occurs in the algebraic formula for the j-invariant of an elliptic
curve. (1728 = 8
2
* 3
3
) The number 1728 is one less than the Hardy–Ramanujan
number 1729 (taxicab number)
The study of this function provides the following representations:
Alternative representations
Series representations
23
Integral representations
24
(1/27((17((6+42.6517+20.3103+((32π)/5)+[((((4(1/2(3/2-euler-mascheroni
constant)+euler-mascheroni constant)π))))]))-e*Pi+Φ)))^2-euler-mascheroni constant
Input interpretation
Result
4095.96… ≈ 4096 = 64
2
((17((6+42.6517+20.3103+((32π)/5)+[((((4(1/2(3/2-euler-mascheroni
constant)+euler-mascheroni constant)π))))]))-7))^1/15
Input interpretation
Result
1.643817466…. ≈ ζ(2) =
2
6
= 1.644934 (trace of the instanton shape)
25
Now, we have that:
From:
For n = 3 and γ = 3/2 = 1.5 , we obtain:
1/(2^1.5*Pi)*gamma(1/2(3-1-1.5))/gamma(1/2*1.5) * Integrate(((1/((x-y)^(3-1-1.5))
u(y))))ℌ^2
Input
Result
The study of this function provides the following representations:
Alternate form
26
Series expansion of the integral at x=0
Indefinite integral assuming all variables are real
From:
we obtain:
0.665936 (-y)^0.5 ℌ^2 u(y) + (0.332968 x ℌ^2 u(y))/(-y)^0.5 - (0.083242 x^2 (ℌ^2
u(y)))/(-y)^1.5 + (0.041621 x^3 ℌ^2 u(y))/(-y)^2.5 - (0.0260131 x^4 (ℌ^2 u(y)))/(-
y)^3.5 + O(x^5)
Input interpretation
27
Result
The study of this function provides the following representations:
Alternate forms
Alternate forms assuming x, y, and ℌ are positive
28
Series expansion at x=0
Series expansion at x=∞
29
Derivative
From the above alternate form
we obtain:
Input interpretation
The study of this function provides the following representations:
Alternate forms
30
Alternate form assuming x, y, and ℌ are positive
Expanded forms
Series expansion at x=0
31
Series expansion at x=∞
Derivative
Now, we have:
Input
Exact result
Decimal form
2.25
From the result:
for x = 1 and y = 2 , we obtain :
0.665936*2.25*2(1-2)^0.5
Input interpretation
32
Result
Polar coordinates
2.99671
From the above alternate form:
we have, for x = 1 and y = 2:
1/((-2)^3.5) ((-2)^3.5-
2.25*2(0.0260131+0.041621*2+0.083242*2^2+0.332968*2^3-0.665936*2^4))
Input interpretation
Result
Polar coordinates
3.16474
33
From:
we obtain:
1/(2^1.5*Pi)*gamma(1/2(3-1-1.5))/gamma(1/2*1.5) * Integrate(((1/((x-y)^(3-1-
1.5)))))ℌ^2
Input
Result
The study of this function provides the following representations:
Alternate form
Series expansion of the integral at x=0
Indefinite integral assuming all variables are real
34
From the above expression
we obtain:
Input interpretation
Result
The study of this function provides the following representations:
Alternate forms
35
Alternate forms assuming x, y, and ℌ are positive
36
Series expansion at x=0
Series expansion at x=∞
Derivative
From the previous alternate form:
we obtain:
37
Input interpretation
The study of this function provides the following representations:
Alternate forms
Alternate form assuming x, y, and ℌ are positive
Expanded forms
38
Series expansion at x=0
Series expansion at x=∞
Derivative
39
From the above alternate form:
for x = 1 and y = 2 , we obtain:
1+1/((-2)^3.5) 2.25((0.0260131-0.041621*2-0.083242*2^2-
0.332968*2^3+0.665936*2^4))
Input interpretation
Result
Polar coordinates
1.81248
From the result
we obtain:
0.665936*2.25(1-2)^0.5
Input interpretation
40
Result
Polar coordinates
1.49836
Dividing the two analyzed expressions, we obtain:
((O(x^5)+(ℌ^2 u(y)(-0.0260131x^4-0.041621x^3y-0.083242x^2y^2-
0.332968xy^3+0.665936y^4))/(-y)^(7/2)))/((O(x^5)+(ℌ^2(-0.0260131x^4-
0.041621x^3y-0.083242x^2y^2-0.332968xy^3+0.665936y^4))/(-y)^(7/2)))
Input interpretation
The study of this function provides the following representations:
Alternate forms
41
Expanded forms
42
Alternate forms assuming x, y, and ℌ are positive
43
Series expansion at x=0
Series expansion at x=∞
Derivative
44
From:
we obtain:
Input interpretation
The study of this function provides the following representations:
Alternate forms
45
Expanded forms
Alternate forms assuming x, y, and ℌ are positive
46
Series expansion at x=0
Series expansion at x=∞
Derivative
47
Subtracting the two above expressions, we obtain:
(ℌ^2 u(y) (-0.0260131 2^4 - 0.041621 2 - 0.083242 2^2 - 0.332968 2^3 + 0.665936
2^4))/(-2)^(7/2) - (ℌ^2 (-0.0260131 - 0.041621 2 - 0.083242 2^2 - 0.332968 2^3 +
0.665936 2^4))/(-2)^(7/2)
Input interpretation
Result
The study of this function provides the following representations:
Alternate forms
Alternate form assuming y and ℌ are real
48
Properties as a real function
Domain
Range
Series expansion at y=0
Derivative
From:
we obtain:
(0.632756 i) (2.25) (2) - (0.667244 i) (2.25)
Input interpretation
Result
49
Polar coordinates
1.3461
Considering only the result of the second integral, we obtain also:
((O(x^5)+(ℌ^2(-0.0260131x^4-0.041621x^3y-0.083242x^2y^2-
0.332968xy^3+0.665936y^4))/(-y)^(7/2)))
Input interpretation
Result
The study of this function provides the following representations:
Alternate forms
50
Alternate form assuming x, y, and ℌ are positive
Expanded forms
51
Series expansion at x=0
Series expansion at x=∞
Derivative
From the previous result:
we obtain, for x = 1 and y =2 :
Input interpretation
52
Result
The study of this function provides the following representations:
Alternate forms
Alternate form assuming ℌ is real
Complex roots
53
Polynomial discriminant
Property as a function
Parity
Derivative
Indefinite integral
From:
(0.632756 i) ℌ^2 + O(1)
we obtain:
(0.632756 i) (2.25) + 1
Input interpretation
54
Result
Polar coordinates
1.73981
From the division:
we obtain, for x = 1 and y = 2:
(38.4422 (-2)^(7/2) - (2.25)(2) (1 + 1.6 2 + 3.2 2^2 + 12.8 2^3 - 25.6 2^4))/(38.4422
(-2)^(7/2) - (2.25) (1 + 1.6 2 + 3.2 2^2 + 12.8 2^3 - 25.6 2^4))
Input interpretation
Result
Polar coordinates
1.75443
55
From:
and
we obtain, after some calculations:
1/(1/4(((38.4422 (-2)^(7/2) - (2.25)(2) (1+1.6*2+3.2*4+12.8*8 - 25.6*16))/(38.4422
(-2)^(7/2) - (2.25) (1 + 1.6*2 + 3.2*4 + 12.8*8 - 25.6*16)))+(((0.632756 i) (2.25) (2)
- (0.667244 i) (2.25)))))
Input interpretation
Result
Polar coordinates
1.6153 result that is a very good approximation to the value of the golden ratio
1.618033988749...
From the two already analyzed expressions:
(0.632756 i) (2.25) (2) - (0.667244 i) (2.25)
Input interpretation
56
Result
Polar coordinates
1.3461
And:
(0.632756 i) ℌ^2 + O(1)
(0.632756 i) (2.25) + 1
Input interpretation
Result
Polar coordinates
1.73981
after some calculations, we obtain:
((2((((1/(2Pi)(((0.632756 i) (2.25) + 1 + ((0.632756 i) (2.25) (2) - (0.667244 i)
(2.25))))^2))))))^(1/(0.5683000031+0.5269391135+0.9568666373))))
where 0.5683000031, 0.5269391135 and 0.9568666373 are the values of the
following Rogers-Ramanujan continued fractions:
57
and
Input interpretation
Result
Polar coordinates
1.64014 ≈ ζ(2) =
2
6
= 1.644934 (trace of the instanton shape)
58
From:
For x = 1 , y = 2, u = 8+4i , n = 3 ; α = 2 , and considering:
Input
Exact result
Decimal form
2.25
From
we obtain:
2* integrate(((1/(x-y)*((8+4i)x-(8+4i)y))))ℌ^2
Indefinite integral
59
The study of this function provides the following representations:
3D plots
Real part (figures that can be related to a D-branes/Instantons)
Imaginary part
60
Contour plots
Real part
Imaginary part
(16 + 8 i) (2.25)
Input
Result
61
Polar coordinates
40.2492
From:
we obtain:
Integrate((((8+4i)*((16 + 8 i)*x (2.25))*(8+4i)))) ℌ^2
Indefinite integral
62
The study of this function provides the following representations:
3D plots
Real part (figures that can be related to a D-branes/Instantons)
Imaginary part
63
Contour plots
Real part
Imaginary part
Alternate form assuming x and ℌ are real
(288 + 1584 i) (2.25)
Input
64
Result
Polar coordinates
3622.43
From the two previous expressions, after some calculations, we obtain:
(1/2((288 + 1584 i) (2.25)) - (1/2((16 + 8 i) (2.25)))i)-64i-Pi*i
Input
Result
Polar coordinates
1729.22
This result is very near to the mass of candidate glueball f
0
(1710) scalar meson.
Furthermore, 1728 occurs in the algebraic formula for the j-invariant of an elliptic
curve. (1728 = 8
2
* 3
3
) The number 1728 is one less than the Hardy–Ramanujan
number 1729 (taxicab number)
65
Polar forms
The study of this function provides the following representations:
Alternative representations
Series representations
66
Integral representations
67
From:
for:
= (16 + 8 i) (2.25)
Input
Result
Polar coordinates
40.2492
μ = 0.665936*2.25(1-2)^0.5
Input interpretation
Result
Polar coordinates
1.49836
68
and for: x = 1 , y = 2, u = 8+4i , n = 3 ; α = 2 , we obtain:
2^2*Pi*gamma(1/2)/gamma(1/2) (((2*((0.665936*2.25(1-2)^0.5))-((((16 + 8 i)
(2.25))^2)))))
Input interpretation
Result
Polar coordinates
20327.4
The study of this function provides the following representations:
Polar forms
69
Alternative representations
Series representations
70
Integral representations
71
From which, we obtain, after some calculations:
(((2^2*Pi*gamma(1/2)/gamma(1/2) (((2*((0.665936*2.25(1-2)^0.5))-((((16 + 8 i)
(2.25))^2))))))))^1/20
Input interpretation
Result
Polar coordinates
1.64212 ≈ ζ(2) =
2
6
= 1.644934 (trace of the instanton shape)
72
From:
AdS cycles in eternally inflating background
Zhi-Guo Liu and Yun-Song Piao - arXiv:1404.5748v1 [hep-th] 23 Apr 2014
We have that:
We consider:
= 4.341 × 10
-9
kg = 2.435 × 10
18
GeV/c
2
.
From the result of the above integral, we obtain:
1/4(3(2.435*10^18)^2)
Input interpretation
Result
Scientific notation
4.44691875*10
36
73
From which:
ln((((3*(2.435*10^18)^2))/4))+18
Input interpretation
Result
102.385274…..
For:
M
P
= 2.435 * 10
18
GeV and H = 2.301 * 10
-18
From:
we obtain:
(0.050^(1.5))/(sqrt(2Pi^2))* x = (2.301*10^-18)/(sqrt(3/2))* (2.435 × 10^18)
Input interpretation
74
Plot
Alternate form
Alternate form assuming x is real
Solution
1817.94
Thence:
(0.050^(1.5))/(sqrt(2Pi^2))* 1817.94
Input interpretation
Result
4.574774392….
75
The study of this function provides the following representations:
Series representations
From:
Considering:
76
from:
we obtain:
2*x*(2.435*10^18)^2 = 3(2.435*10^18)^2
Input interpretation
Result
Plot
Alternate form
Alternate form assuming x is real
Solution
x = 3/2 =
77
Thence:
2*3/2*(2.435*10^18)^2
Input interpretation
Result
Scientific notation
1.7787675*10
37
From:
Sqrt(2*(2.435*10^18)^2*3/2)*1817.94
Input interpretation
Result
7.66724…*10
21
78
Furthermore, from:
we obtain:
((Sqrt(3/2)*(2.435*10^18)))/(2.301*10^-18)
Input interpretation
Result
1.29607…*10
36
From:
and
we obtain also:
1/2(0.9991104684+0.9568666373)+1/((((1/4(3(2.435*10^18)^2)) /
(((((Sqrt(3/2)*(2.435*10^18)))/(2.301*10^-18))))))^1/3)
where 0.9991104684 and 0.9568666373 are the values of the following Rogers-
Ramanujan continued fractions:
79
Input interpretation
Result
1.641002901498….≈ ζ(2) =
2
6
= 1.644934 (trace of the instanton shape)
From the two expressions:
and
we obtain:
80
((2*3/2*(2.435*10^18)^2)/(1/4(3(2.435*10^18)^2)))
Input interpretation
Result
4
and after some easy calculations:
((2*3/2*(2.435*10^18)^2)/(1/4(3(2.435*10^18)^2)))^3
Input interpretation
Result
64 = 8
2
From which:
27((2*3/2*(2.435*10^18)^2)/(1/4(3(2.435*10^18)^2)))^3+1
Input interpretation
81
Result
1729
This result is very near to the mass of candidate glueball f
0
(1710) scalar meson.
Furthermore, 1728 occurs in the algebraic formula for the j-invariant of an elliptic
curve. (1728 = 8
2
* 3
3
) The number 1728 is one less than the Hardy–Ramanujan
number 1729 (taxicab number)
(27((2*3/2*(2.435*10^18)^2)/(1/4(3(2.435*10^18)^2)))^3+1)^1/15
Input interpretation
Result
1.64381522874….≈ ζ(2) =
2
6
= 1.644934 (trace of the instanton shape)
(((2*3/2*(2.435*10^18)^2)/(1/4(3(2.435*10^18)^2)))^3)^2
Input interpretation
Result
4096 = 64
2
82
From:
and
we obtain:
(2*3/2*(2.435*10^18)^2) 1/(((Sqrt(3/2)*(2.435*10^18)))/(2.301*10^-18))
Input interpretation
Result
13.72433181198….
and dividing also from:
we obtain :
(2*3/2*(2.435*10^18)^2) 1/(((Sqrt(3/2)*(2.435*10^18)))/(2.301*10^-18))
*1/((((0.050^(1.5))/(sqrt(2Pi^2))* 1817.94)))
83
Input interpretation
Result
3
Multiplying by the previous expression
we obtain also:
((2*3/2*(2.435*10^18)^2)/(1/4(3(2.435*10^18)^2)))(2*3/2*(2.435*10^18)^2)
1/(((Sqrt(3/2)*(2.435*10^18)))/(2.301*10^-18)) *1/((((0.050^(1.5))/(sqrt(2Pi^2))*
1817.94)))
Input interpretation
Result
12
84
From which:
((((2*3/2*(2.435*10^18)^2)/(1/4(3(2.435*10^18)^2)))(2*3/2*(2.435*10^18)^2)
1/(((Sqrt(3/2)*(2.435*10^18)))/(2.301*10^-18)) *1/((((0.050^(1.5))/(sqrt(2Pi^2))*
1817.94)))))^3+1
Input interpretation
Result
1729
This result is very near to the mass of candidate glueball f
0
(1710) scalar meson.
Furthermore, 1728 occurs in the algebraic formula for the j-invariant of an elliptic
curve. (1728 = 8
2
* 3
3
) The number 1728 is one less than the Hardy–Ramanujan
number 1729 (taxicab number)
85
(((((2*3/2*(2.435*10^18)^2)/(1/4(3(2.435*10^18)^2)))(2*3/2*(2.435*10^18)^2)
1/(((Sqrt(3/2)*(2.435*10^18)))/(2.301*10^-18)) *1/((((0.050^(1.5))/(sqrt(2Pi^2))*
1817.94)))))^3+1)^1/15
Input interpretation
Result
1.64382….≈ ζ(2) =
2
6
= 1.644934 (trace of the instanton shape)
86
(1/27(((((2*3/2*(2.435*10^18)^2)/(1/4(3(2.435*10^18)^2)))(2*3/2*(2.435*10^18)^2
) 1/(((Sqrt(3/2)*(2.435*10^18)))/(2.301*10^-18)) *1/((((0.050^(1.5))/(sqrt(2Pi^2))*
1817.94)))))^3))^2
Input interpretation
Result
4096.02….≈ 4096 = 64
2
87
Observations
We note that, from the number 8, we obtain as follows:
We notice how from the numbers 8 and 2 we get 64, 1024, 4096 and 8192, and that 8
is the fundamental number. In fact 8
2
= 64, 8
3
= 512, 8
4
= 4096. We define it
"fundamental number", since 8 is a Fibonacci number, which by rule, divided by the
previous one, which is 5, gives 1.6 , a value that tends to the golden ratio, as for all
numbers in the Fibonacci sequence
88
“Golden” Range
Finally we note how 8
2
= 64, multiplied by 27, to which we add 1, is equal to 1729,
the so-called "Hardy-Ramanujan number". Then taking the 15th root of 1729, we
obtain a value close to ζ(2) that 1.6438 ..., which, in turn, is included in the range of
what we call "golden numbers"
Furthermore for all the results very near to 1728 or 1729, adding 64 = 8
2
, one obtain
values about equal to 1792 or 1793. These are values almost equal to the Planck
multipole spectrum frequency 1792.35 and to the hypothetical Gluino mass
89
Appendix
From: A. Sagnotti – AstronomiAmo, 23.04.2020
In the above figure, it is said that: “why a given shape of the extra dimensions?
Crucial, it determines the predictions for α”.
We propose that whatever shape the compactified dimensions are, their geometry
must be based on the values of the golden ratio and ζ(2), (the latter connected to 1728
or 1729, whose fifteenth root provides an excellent approximation to the above
mentioned value) which are recurrent as solutions of the equations that we are going
to develop. It is important to specify that the initial conditions are always values
belonging to a fundamental chapter of the work of S. Ramanujan "Modular equations
and Appoximations to Pi" (see references). These values are some multiples of 8 (64
and 4096), 276, which added to 4096, is equal to 4372, and finally e
π√22
90
We have, in certain cases, the following connections:
Fig. 1
Fig. 2
91
Fig. 3
Stringscape - a small part of the string-theory landscape showing the new de Sitter solution as a local
minimum of the energy (vertical axis). The global minimum occurs at the infinite size of the extra
dimensions on the extreme right of the figure.
Fig. 4
92
With regard the Fig. 4 the points of arrival and departure on the right-hand side of the
picture are equally spaced and given by the following equation:
we obtain:
2Pi/(ln(2))
Input:
Exact result:
Decimal approximation:
9.06472028365….
Alternative representations:
93
Series representations:
Integral representations:
94
From which:
(2Pi/(ln(2)))*(1/12 π log(2))
Input:
Exact result:
Decimal approximation:
1.6449340668…. = ζ(2) =
2
6
= 1.644934
95
From:
Modular equations and approximations to - Srinivasa Ramanujan
Quarterly Journal of Mathematics, XLV, 1914, 350 – 372
We have that:
96
We note that, with regard 4372, we can to obtain the following results:
27((4372)^1/2-2-1/2(((√(10-2√5) -2))⁄((√5-1))))+φ
Input
Result
Decimal approximation
1729.0526944….
This result is very near to the mass of candidate glueball f
0
(1710) scalar meson.
Furthermore, 1728 occurs in the algebraic formula for the j-invariant of an elliptic
curve. (1728 = 8
2
* 3
3
) The number 1728 is one less than the Hardy–Ramanujan
number 1729 (taxicab number)
Alternate forms
97
Minimal polynomial
Expanded forms
Series representations
98
Or:
27((4096+276)^1/2-2-1/2(((√(10-2√5) -2))⁄((√5-1))))+φ
99
Input
Result
Decimal approximation
1729.0526944…. as above
Alternate forms
100
Minimal polynomial
Expanded forms
Series representations
101
102
From which:
(27((4372)^1/2-2-1/2(((√(10-2√5) -2))⁄((√5-1))))+φ)^1/15
Input
Exact result
Decimal approximation
1.64381856858…. ≈ ζ(2) =
2
6
= 1.644934
Alternate forms
103
Minimal polynomial
Expanded forms
All 15th roots of ϕ + 27 (-2 + 2 sqrt(1093) - (sqrt(10 - 2 sqrt(5)) - 2)/(2 (sqrt(5) -
1)))
104
Series representations
105
Integral representation
106
From:
An Update on Brane Supersymmetry Breaking
J. Mourad and A. Sagnotti - arXiv:1711.11494v1 [hep-th] 30 Nov 2017
From the following vacuum equations:
we have obtained, from the results almost equals of the equations, putting
instead of
a new possible mathematical connection between the two exponentials. Thence, also
the values concerning p, C, β
E
and correspond to the exponents of e (i.e. of exp).
Thence we obtain for p = 5 and β
E
= 1/2:
6+
= 4096
18
Therefore, with respect to the exponentials of the vacuum equations, the Ramanujan’s
exponential has a coefficient of 4096 which is equal to 64
2
, while -6C+ is equal to -
18. From this it follows that it is possible to establish mathematically, the dilaton
value.
107
For
exp((-Pi*sqrt(18)) we obtain:
Input:
Exact result:
Decimal approximation:
1.6272016… * 10
-6
Property:
Series representations:
108
Now, we have the following calculations:
6+
= 4096
18
18
= 1.6272016… * 10^-6
from which:
1
4096
6+
= 1.6272016… * 10^-6
0.000244140625
6+
=
18
= 1.6272016… * 10^-6
Now:
ln
18
= 13.328648814475 =
18
And:
(1.6272016* 10^-6) *1/ (0.000244140625)
Input interpretation:
Result:
0.006665017...
109
Thence:
0.000244140625
6+
=
18
Dividing both sides by 0.000244140625, we obtain:
0.000244140625
0.000244140625
6+
=
1
0.000244140625
18
6+
= 0.0066650177536
((((exp((-Pi*sqrt(18)))))))*1/0.000244140625
Input interpretation:
Result:
0.00666501785…
Series representations:
110
Now:
6+
= 0.0066650177536
=
= 0.00666501785…
From:
ln(0.00666501784619)
Input interpretation:
Result:
-5.010882647757…
111
Alternative representations:
Series representations:
Integral representation:
In conclusion:
6 + = 5.010882647757
and for C = 1, we obtain:
112
= 5.010882647757 + 6 = . =
Note that the values of n
s
(spectral index) 0.965, of the average of the Omega mesons
Regge slope 0.987428571 and of the dilaton 0.989117352243, are also connected to
the following two Rogers-Ramanujan continued fractions:
(http://www.bitman.name/math/article/102/109/)
Also performing the 512
th
root of the inverse value of the Pion meson rest mass
139.57, we obtain:
((1/(139.57)))^1/512
Input interpretation:
113
Result:
0.99040073.... result very near to the dilaton value . = and to
the value of the following Rogers-Ramanujan continued fraction:
From
Properties of Nilpotent Supergravity
E. Dudas, S. Ferrara, A. Kehagias and A. Sagnotti - arXiv:1507.07842v2 [hep-th] 14
Sep 2015
We have that:
We analyzing the following equation:
114
We have:
(M^2)/3*[1-(b/euler number * k/sqrt6) * (φ- sqrt6/k) * exp(-(k/sqrt6)(φ- sqrt6/k))]^2
i.e.
V = (M^2)/3*[1-(b/euler number * k/sqrt6) * (φ- sqrt6/k) * exp(-(k/sqrt6)(φ-
sqrt6/k))]^2
For k = 2 and φ = 0.9991104684, that is the value of the scalar field that is equal to
the value of the following Rogers-Ramanujan continued fraction:
we obtain:
V = (M^2)/3*[1-(b/euler number * 2/sqrt6) * (0.9991104684- sqrt6/2) * exp(-
(2/sqrt6)(0.9991104684- sqrt6/2))]^2
Input interpretation:
115
Result:
Solutions:
Alternate forms:
Expanded form:
Alternate form assuming b, M, and V are positive:
Alternate form assuming b, M, and V are real:
116
Derivative:
Implicit derivatives
117
Global minimum:
Global minima:
From:
we obtain:
(225.913 (-0.054323 M^2 + 6.58545×10^-10 sqrt(M^4)))/M^2
Input interpretation:
118
Result:
Plots:
Alternate form assuming M is real:
-12.2723 result very near to the black hole entropy value 12.1904 = ln(196884)
Alternate forms:
119
Expanded form:
Property as a function:
Parity
Series expansion at M = 0:
Series expansion at M = ∞:
Derivative:
120
Indefinite integral:
Global maximum:
Global minimum:
Limit:
121
Definite integral after subtraction of diverging parts:
From b that is equal to
From:
we obtain:
1/3 (0.0814845 ((225.913 (-0.054323 M^2 + 6.58545×10^-10 sqrt(M^4)))/M^2 ) +
1)^2 M^2
Input interpretation:
Result:
122
Plots: (possible mathematical connection with an open string)
M = -0.5; M = 0.2
(possible mathematical connection with an open string)
M = 2 ; M = 3
Root:
Property as a function:
Parity
123
Series expansion at M = 0:
Series expansion at M = ∞:
Definite integral after subtraction of diverging parts:
For M = - 0.5 , we obtain:
124
1/3 (0.0814845 ((225.913 (-0.054323 (-0.5)^2 + 6.58545×10^-10 sqrt((-0.5)^4)))/(-
0.5)^2 ) + 1)^2 * (-0.5^2)
Input interpretation:
Result:
-4.38851344947*10
-16
For M = 0.2:
1/3 (0.0814845 ((225.913 (-0.054323 0.2^2 + 6.58545×10^-10 sqrt(0.2^4)))/0.2^2 ) +
1)^2 0.2^2
Input interpretation:
125
Result:
7.021621519159*10
-17
For M = 3:
1/3 (0.0814845 ((225.913 (-0.054323 3^2 + 6.58545×10^-10 sqrt(3^4)))/3^2 ) + 1)^2
3^2
Input interpretation:
Result:
1.57986484181*10
-14
126
For M = 2:
1/3 (0.0814845 ((225.913 (-0.054323 2^2 + 6.58545×10^-10 sqrt(2^4)))/2^2 ) + 1)^2
2^2
Input interpretation:
Result:
7.021621519*10
-15
From the four results
7.021621519*10^-15 ; 1.57986484181*10^-14 ; 7.021621519159*10^-17 ;
-4.38851344947*10^-16
we obtain, after some calculations:
sqrt[1/(2Pi)(7.021621519*10^-15 + 1.57986484181*10^-14 +7.021621519*10^-17 -
4.38851344947*10^-16)]
127
Input interpretation:
Result:
5.9776991059*10
-8
result very near to the Planck's electric flow 5.975498 × 10
−8
that
is equal to the following formula:
We note that:
1/55*(([(((1/[(7.021621519*10^-15 + 1.57986484181*10^-14 +7.021621519*10^-17
-4.38851344947*10^-16)])))^1/7]-((log^(5/8)(2))/(2 2^(1/8) 3^(1/4) e log^(3/2)(3)))))
Input interpretation:
Result:
1.6181818182… result that is a very good approximation to the value of the golden
ratio 1.618033988749...
128
From the Planck units:
Planck Length
5.729475 * 10
-35
Lorentz-Heaviside value
Planck’s Electric field strength
1.820306 * 10
61
V*m Lorentz-Heaviside value
Planck’s Electric flux
5.975498*10
-8
V*m Lorentz-Heaviside value
Planck’s Electric potential
1.042940*10
27
V Lorentz-Heaviside value
129
Relationship between Planck’s Electric Flux and Planck’s Electric Potential
E
P
* l
P
= (1.820306 * 10
61
) * 5.729475 * 10
-35
Input interpretation:
Result:
Scientific notation:
1.042939771935*10
27
≈ 1.042940*10
27
Or:
E
P
* l
P
2
/ l
P
= (5.975498*10
-8
)*1/(5.729475 * 10
-35
)
Input interpretation:
Result:
1.042939885417*10
27
≈ 1.042940*10
27
130
Acknowledgments
M. Nardelli thanks Francesco Maggi, Professor of Mathematics at University of
Texas - Austin, Department of Mathematics, for his availability and kindness towards
him
131
References
Isoperimetry and Stability Properties of Balls with Respect to Nonlocal Energies
A. Figalli, N. Fusco, F. Maggi, V. Millot, M. Morini - Commun. Math. Phys Digital
Object Identifier (DOI) 10.1007/s00220-014-2244-1
AdS cycles in eternally inflating background
Zhi-Guo Liu and Yun-Song Piao - arXiv:1404.5748v1 [hep-th] 23 Apr 2014
Modular equations and approximations to - Srinivasa Ramanujan
Quarterly Journal of Mathematics, XLV, 1914, 350 – 372
An Update on Brane Supersymmetry Breaking
J. Mourad and A. Sagnotti - arXiv:1711.11494v1 [hep-th] 30 Nov 2017
Properties of Nilpotent Supergravity
E. Dudas, S. Ferrara, A. Kehagias and A. Sagnotti - arXiv:1507.07842v2 [hep-th] 14
Sep 2015